Problem: Solve the congruence $11n \equiv 7 \pmod{43}$, as a residue modulo 43.  (Give an answer between 0 and 42.)
Solution: Note that 43 is a close to a multiple of 11, namely 44.  Multiplying both sides of the given congruence by 4, we get $44n \equiv 28 \pmod{43}$, which reduces to $n \equiv \boxed{28} \pmod{43}$.